If $\lim\limits_{x\rightarrow x_0} f(x) = \infty$ then $\exists \alpha<x_0\exists C\ :\ 1+f^2(x)\le Cf^2(x)\ \forall x\in[\alpha,x_0)$

Question

Why is the following true:

If $f:[0,x_0)\mapsto\Bbb R$ and $\lim\limits_{x\rightarrow x_0} f(x) = \infty$ then $\exists \alpha<x_0\exists C>0\ :\ 1+f^2(x)\le Cf^2(x)\ \forall x\in[\alpha,x_0)$

I dont see why this property should hold. If not true, we have $\forall \alpha,C\ \exists x\in[\alpha,x_0)\ :\ 1+f^2(x)> Cf^2(x)$

So for example if $\alpha = x_0-1/n$ and $C=n$. Let $(x_n)$ be a sequence such that $x_0-1/n\le x_n <x_0$. Clearly $\lim x_n = x_0$

And for each $n$ the following holds:

$1>(n-1)f^2(x_n)\iff{1\over n-1}>f^2(x_n)\implies \lim f(x_n)=0$ in particular $\lim f<\infty$

Does this show the implication?

Answer 1

Since $\lim\limits_{x\rightarrow x_0} f(x)=\infty$ then, by definition of limit, $\exists \alpha<x_0$ such that $f^2(x)>1$ $\forall x\in[\alpha,x_0)$. Then $$1+f^2(x)\le Cf^2(x)\quad \forall x\in[\alpha,x_0)$$ trivially holds by letting $C\geq 2$. Actually the same conclusion holds by assuming the weaker condition $\lim\limits_{x\rightarrow x_0} |f(x)|>1.$